3.25.79 \(\int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+(-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+(20+5 e^2+5 e^{10}-10 x) \log (x)) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx\)

Optimal. Leaf size=22 \[ 5 \left (4+e^2+e^{10}-x\right ) x \log (5+x-\log (x)) \]

________________________________________________________________________________________

Rubi [F]  time = 0.51, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20 + E^2*(5 - 5*x) + E^10*(5 - 5*x) - 25*x + 5*x^2 + (-100 + E^2*(-25 - 5*x) + E^10*(-25 - 5*x) + 30*x +
10*x^2 + (20 + 5*E^2 + 5*E^10 - 10*x)*Log[x])*Log[5 + x - Log[x]])/(-5 - x + Log[x]),x]

[Out]

-5*(4 + E^2 + E^10)*Defer[Int][(5 + x - Log[x])^(-1), x] + 5*(5 + E^2 + E^10)*Defer[Int][x/(5 + x - Log[x]), x
] - 5*Defer[Int][x^2/(5 + x - Log[x]), x] + 5*(4 + E^2 + E^10)*Defer[Int][Log[5 + x - Log[x]], x] - 10*Defer[I
nt][x*Log[5 + x - Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20+\left (e^2+e^{10}\right ) (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx\\ &=\int \frac {5 \left (-4+e^2 \left (1+e^8\right ) (-1+x)+5 x-x^2+\left (4+e^2+e^{10}-2 x\right ) (5+x-\log (x)) \log (5+x-\log (x))\right )}{5+x-\log (x)} \, dx\\ &=5 \int \frac {-4+e^2 \left (1+e^8\right ) (-1+x)+5 x-x^2+\left (4+e^2+e^{10}-2 x\right ) (5+x-\log (x)) \log (5+x-\log (x))}{5+x-\log (x)} \, dx\\ &=5 \int \left (\frac {-4-e^2-e^{10}+\left (5+e^2+e^{10}\right ) x-x^2}{5+x-\log (x)}+\left (4+e^2+e^{10}-2 x\right ) \log (5+x-\log (x))\right ) \, dx\\ &=5 \int \frac {-4-e^2-e^{10}+\left (5+e^2+e^{10}\right ) x-x^2}{5+x-\log (x)} \, dx+5 \int \left (4+e^2+e^{10}-2 x\right ) \log (5+x-\log (x)) \, dx\\ &=5 \int \left (-\frac {4 \left (1+\frac {1}{4} e^2 \left (1+e^8\right )\right )}{5+x-\log (x)}+\frac {\left (5+e^2+e^{10}\right ) x}{5+x-\log (x)}-\frac {x^2}{5+x-\log (x)}\right ) \, dx+5 \int \left (4 \left (1+\frac {1}{4} e^2 \left (1+e^8\right )\right ) \log (5+x-\log (x))-2 x \log (5+x-\log (x))\right ) \, dx\\ &=-\left (5 \int \frac {x^2}{5+x-\log (x)} \, dx\right )-10 \int x \log (5+x-\log (x)) \, dx-\left (5 \left (4+e^2+e^{10}\right )\right ) \int \frac {1}{5+x-\log (x)} \, dx+\left (5 \left (4+e^2+e^{10}\right )\right ) \int \log (5+x-\log (x)) \, dx+\left (5 \left (5+e^2+e^{10}\right )\right ) \int \frac {x}{5+x-\log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 22, normalized size = 1.00 \begin {gather*} 5 \left (4+e^2+e^{10}-x\right ) x \log (5+x-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + E^2*(5 - 5*x) + E^10*(5 - 5*x) - 25*x + 5*x^2 + (-100 + E^2*(-25 - 5*x) + E^10*(-25 - 5*x) + 3
0*x + 10*x^2 + (20 + 5*E^2 + 5*E^10 - 10*x)*Log[x])*Log[5 + x - Log[x]])/(-5 - x + Log[x]),x]

[Out]

5*(4 + E^2 + E^10 - x)*x*Log[5 + x - Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 27, normalized size = 1.23 \begin {gather*} -5 \, {\left (x^{2} - x e^{10} - x e^{2} - 4 \, x\right )} \log \left (x - \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x-25)*exp(2)+10*x^2+30*x-100)*log(x-lo
g(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp(2)+5*x^2-25*x+20)/(log(x)-5-x),x, algorithm="fricas")

[Out]

-5*(x^2 - x*e^10 - x*e^2 - 4*x)*log(x - log(x) + 5)

________________________________________________________________________________________

giac [B]  time = 0.37, size = 51, normalized size = 2.32 \begin {gather*} -5 \, x^{2} \log \left (x - \log \relax (x) + 5\right ) + 5 \, x e^{10} \log \left (x - \log \relax (x) + 5\right ) + 5 \, x e^{2} \log \left (x - \log \relax (x) + 5\right ) + 20 \, x \log \left (x - \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x-25)*exp(2)+10*x^2+30*x-100)*log(x-lo
g(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp(2)+5*x^2-25*x+20)/(log(x)-5-x),x, algorithm="giac")

[Out]

-5*x^2*log(x - log(x) + 5) + 5*x*e^10*log(x - log(x) + 5) + 5*x*e^2*log(x - log(x) + 5) + 20*x*log(x - log(x)
+ 5)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 29, normalized size = 1.32

method	result	size
risch	\(\left (5 x \,{\mathrm e}^{10}+5 \,{\mathrm e}^{2} x -5 x^{2}+20 x \right ) \ln \left (x -\ln \relax (x )+5\right )\)	\(29\)
norman	\(\left (5 \,{\mathrm e}^{10}+5 \,{\mathrm e}^{2}+20\right ) x \ln \left (x -\ln \relax (x )+5\right )-5 \ln \left (x -\ln \relax (x )+5\right ) x^{2}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*exp(5)^2+5*exp(2)+20-10*x)*ln(x)+(-5*x-25)*exp(5)^2+(-5*x-25)*exp(2)+10*x^2+30*x-100)*ln(x-ln(x)+5)+(
-5*x+5)*exp(5)^2+(-5*x+5)*exp(2)+5*x^2-25*x+20)/(ln(x)-5-x),x,method=_RETURNVERBOSE)

[Out]

(5*x*exp(10)+5*exp(2)*x-5*x^2+20*x)*ln(x-ln(x)+5)

________________________________________________________________________________________

maxima [A]  time = 0.96, size = 23, normalized size = 1.05 \begin {gather*} -5 \, {\left (x^{2} - x {\left (e^{10} + e^{2} + 4\right )}\right )} \log \left (x - \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x-25)*exp(2)+10*x^2+30*x-100)*log(x-lo
g(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp(2)+5*x^2-25*x+20)/(log(x)-5-x),x, algorithm="maxima")

[Out]

-5*(x^2 - x*(e^10 + e^2 + 4))*log(x - log(x) + 5)

________________________________________________________________________________________

mupad [B]  time = 1.68, size = 20, normalized size = 0.91 \begin {gather*} 5\,x\,\ln \left (x-\ln \relax (x)+5\right )\,\left ({\mathrm {e}}^2-x+{\mathrm {e}}^{10}+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x - log(x - log(x) + 5)*(30*x + log(x)*(5*exp(2) - 10*x + 5*exp(10) + 20) + 10*x^2 - exp(2)*(5*x + 25)
- exp(10)*(5*x + 25) - 100) - 5*x^2 + exp(2)*(5*x - 5) + exp(10)*(5*x - 5) - 20)/(x - log(x) + 5),x)

[Out]

5*x*log(x - log(x) + 5)*(exp(2) - x + exp(10) + 4)

________________________________________________________________________________________

sympy [A]  time = 0.51, size = 29, normalized size = 1.32 \begin {gather*} \left (- 5 x^{2} + 20 x + 5 x e^{2} + 5 x e^{10}\right ) \log {\left (x - \log {\relax (x )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*exp(5)**2+5*exp(2)+20-10*x)*ln(x)+(-5*x-25)*exp(5)**2+(-5*x-25)*exp(2)+10*x**2+30*x-100)*ln(x-l
n(x)+5)+(-5*x+5)*exp(5)**2+(-5*x+5)*exp(2)+5*x**2-25*x+20)/(ln(x)-5-x),x)

[Out]

(-5*x**2 + 20*x + 5*x*exp(2) + 5*x*exp(10))*log(x - log(x) + 5)

________________________________________________________________________________________